In flat spacetime

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Relativity & Gravitation · Differential geometry of curved spacetime

The Geometry of Gravity, Ch Matter in curved spacetime: minimal coupling and relativistic hydrodynamics

Problem

In flat spacetime, in Lorenz gauge (∂μAμ=0\partial_\mu A^\mu=0), Maxwell's equations reduce to ∂α∂αAμ=−μ0Jμ,\partial^\alpha\partial_\alpha A^\mu = -\mu_0 J^\mu, and this is equivalent there to the gauge-invariant form ∂νFμν=μ0Jμ\partial_\nu F^{\mu\nu}=\mu_0 J^\mu with Fμν=∂μAν−∂νAμF^{\mu\nu}=\partial^\mu A^\nu-\partial^\nu A^\mu. Someone argues: "Minimal coupling is a fixed, mechanical rule: replace ημν\eta^{\mu\nu} by gμνg^{\mu\nu} and ∂μ\partial_\mu by ∇μ\nabla_\mu. Applying it to the Lorenz-gauge equation above gives gαβ∇α∇βAμ=−μ0Jμ,g^{\alpha\beta}\nabla_\alpha\nabla_\beta A^\mu = -\mu_0 J^\mu, and because the flat-spacetime Lorenz-gauge equation and the FF-based Maxwell equation were the same equation in flat spacetime, minimally coupling either one must give the same curved-spacetime law. So this is the one and only curved-spacetime Maxwell equation that minimal coupling permits." ((a)) Minimally couple the FF-based Maxwell equation instead (keeping the Lorenz gauge condition ∇μAμ=0\nabla_\mu A^\mu=0 in curved spacetime), and use the commutator identity ∇μ∇νAμ−∇ν∇μAμ=RνμAμ\nabla_\mu\nabla^\nu A^\mu - \nabla^\nu\nabla_\mu A^\mu = R^\nu{}_\mu A^\mu to express the result as □Aμ\Box A^\mu plus a curvature term, where □Aμ≡gαβ∇α∇βAμ\Box A^\mu \equiv g^{\alpha\beta}\nabla_\alpha\nabla_\beta A^\mu. ((b)) Compare this with the claimed unique law above, and say exactly where the argument fails. ((c)) State which of the two forms is the physically correct curved-spacetime Maxwell equation, and why minimal coupling by itself cannot be the reason for preferring it.

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